Asked by Nikita
How do you find factors of this with long division?
2x^3 - 3x^2 + 3x -1 = 0
I just don't know how to do it because there is a coefficient in front of first term.
2x^3 - 3x^2 + 3x -1 = 0
I just don't know how to do it because there is a coefficient in front of first term.
Answers
Answered by
Reiny
After you tried ± 1 and that didn't work, you would now try ± 1/2
sure enough, x = 1/2 works
(2(1/2)^3 - 3(1/2)^2 + 3(1/2) - 1
= 1/4 - 3/4 + 3/2 - 1
= 0
so (2x - 1) is a factor
Now you can do your long division, and get the other factor to be
(x^2 - x + 1)
sure enough, x = 1/2 works
(2(1/2)^3 - 3(1/2)^2 + 3(1/2) - 1
= 1/4 - 3/4 + 3/2 - 1
= 0
so (2x - 1) is a factor
Now you can do your long division, and get the other factor to be
(x^2 - x + 1)
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